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True [87]
3 years ago
15

Let mAngleA = 40°. If AngleB is a complement of AngleA, and AngleC is a supplement of AngleB, find these measures.

Mathematics
2 answers:
ludmilkaskok [199]3 years ago
7 0

Answer:

m\angle B = 50^\circ

m\angle C = 130^\circ

Step-by-step explanation:

We are given the following information in the question:

m\angle A = 40^\circ

Angle B is a complement of Angle A. Thus, we can write:

\angle A + \angle B = 90^\circ\\\angle B = 90 - m\angle A\\\angle B = 90 - 40 = 50\\m\angle B = 50^\circ

Angle C is a supplement of Angle B. Thus, we can write:

\angle C + \angle B = 180^\circ\\\angle C = 180 - m\angle B\\\angle C = 180 - 50 = 130\\m\angle C = 130^\circ

wlad13 [49]3 years ago
5 0

Answer:

the answer is m angle B = 50

m angle C = 130

Step-by-step explanation:

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The required value after simplification of the s = -16/3. None of these are correct.

Given that,
To simplify [\frac{x^{2/3}x^{-1/2}}{x\sqrt{x^3}\sqrt[3]{x}}]^2   and to find the value of s in x^s.

<h3>What is simplification?</h3>

The process in mathematics to operate and interpret the function to make the function simple or more understandable is called simplifying and the process is called simplification.

Simplification,

=[\frac{x^{2/3}x^{-1/2}}{x\sqrt{x^3}\sqrt[3]{x}}]^2\\= \frac{x^{4/3}x^{-1}}{x^2x^3*{x}^{2/3}}\\= \frac{x^{1/3}}{x^{17/3}}\\=x^{-16/3}
Comparing with x^S
s = -16/3


Thus, the required value of the s = -16/3. None of these are correct.

Learn more about simplification here: brainly.com/question/12501526

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1 year ago
Considering only the values of β for which sinβtanβsecβcotβ is defined, which of the following expressions is equivalent to sinβ
-Dominant- [34]

Answer:

\tan(\beta)

Step-by-step explanation:

For many of these identities, it is helpful to convert everything to sine and cosine, see what cancels, and then work to build out to something.  If you have options that you're building toward, aim toward one of them.

{\tan(\theta)}={\dfrac{\sin(\theta)}{\cos(\theta)}    and   {\sec(\theta)}={\dfrac{1}{\cos(\theta)}

Recall the following reciprocal identity:

\cot(\theta)=\dfrac{1}{\tan(\theta)}=\dfrac{1}{ \left ( \dfrac{\sin(\theta)}{\cos(\theta)} \right )} =\dfrac{\cos(\theta)}{\sin(\theta)}

So, the original expression can be written in terms of only sines and cosines:

\sin(\beta)\tan(\beta)\sec(\beta)\cot(\beta)

\sin(\beta) * \dfrac{\sin(\beta) }{\cos(\beta) } * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) } {\sin(\beta) }

\sin(\beta) * \dfrac{\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} {\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}

\sin(\beta) *\dfrac{1 }{\cos(\beta) }

\dfrac{\sin(\beta)}{\cos(\beta) }

Working toward one of the answers provided, this is the tangent function.


The one caveat is that the original expression also was undefined for values of beta that caused the sine function to be zero, whereas this simplified function is only undefined for values of beta where the cosine is equal to zero.  However, the questions states that we are only considering values for which the original expression is defined, so, excluding those values of beta, the original expression is equivalent to \tan(\beta).

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